Question

If kf for water is 1.86 k kg mol–1, what is the freezing point of 0.1 molal solution of a substance which undergoes no dissociation or association of solute

Answer 1

Answer : The freezing point of 0.1 molal solution of a substance is 272.814 K.

Solution : Given,

$k_f=1.86Kkg/mole$

Molality = 0.1 molal

The formula used for depression in freezing point is,

$\Delta T_f=k_f\times m\\T^o_f-T_f=k_f\times m$

where,

$T^o_f$ = freezing point of pure solvent = 273 K

$T_f$ = freezing point of solution

m = molality

$k_f$ = molal depression constant

Now put all the given values in this formula, we get

$273K-T_f=1.86Kkg/mole\times 0.1mole/kg$

$T_f=272.814K$

Therefore, the freezing point of 0.1 molal solution of a substance is 272.814 K.

Answer 2

The freezing point of the given solution would be 272.814 K.

Explanation:

We know ΔTf = i Kfm∆Tf = i Kfm

m = 0.1 m and Kf = 1.86 K Kg mol-1

On substituting the values in above formula we get:

∆Tf  = 1.86 ×× 0.1  

∆Tf  = 0.186 K  

As ​ ∆Tf  = T initial - T final

T initial = 273 K as water freezes at this temperature.

T final  = 272.814 K  

Hence, the freezing point of the given solution would be 272.814 K.